Problem: What is the value of $\dfrac{d}{dx}\left(\dfrac{1}{x^4}-\dfrac{3}{x^2}\right)$ at $x=1$ ?
The strategy We can first rewrite each rational term in the expression as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have the derivative, we can evaluate it at $x=1$. Rewriting rational terms as negative powers $\begin{aligned} &\phantom{=}\dfrac{1}{x^4}-\dfrac{3}{x^2} \\\\ &=x^{-4}-3x^{-2} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(x^{-4}-3x^{-2} ) \\\\ &=\dfrac{d}{dx}(x^{-4})-3\dfrac{d}{dx}(x^{-2}) \\\\ &=-4x^{-5}-3(-2)x^{-3} \\\\ &=-4x^{-5}+6x^{-3} \\\\ &=-\dfrac{4}{x^5}+\dfrac{6}{x^3} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(\dfrac{1}{x^4}-\dfrac{3}{x^2}\right)=-\dfrac{4}{x^5}+\dfrac{6}{x^3}$. Let's evaluate it at $x=1$ : $\begin{aligned} &\phantom{=}-\dfrac{4}{x^5}+\dfrac{6}{x^3} \\\\ &=-\dfrac{4}{(1)^5}+\dfrac{6}{(1)^3} \\\\ &=-4+6 \\\\ &=2 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\dfrac{1}{x^4}-\dfrac{3}{x^2}\right)$ at $x=1$ is $2$.